$\cos x \cdot \cos 7 x-\cos 5 x \cdot \cos 13 x=$
$\cos x \cdot \cos 7 x-\cos 5 x \cdot \cos 13 x=$
- $2 \cos ^{2} 6 x \cdot \cos 12 x$
- $2 \sin ^{2} 6 x \cdot \cos 6 x$
- $2 \sin 6 x \cdot \sin 12 x$
- $2 \sin 6 x \cdot \cos 12 x$
Solution
$\cos x \cos 7 x-\cos 5 x \cos 13 x$
$\begin{array}{l}
=\frac{1}{2}(2 \cos x \cos 7 x-2 \cos 5 x \cos 13 x) \\
=\frac{1}{2}(\cos 8 x+\cos 6 x-\cos 18 x-\cos 8 x) \\
=\frac{1}{2}(\cos 6 x-\cos 18 x)=\frac{1}{2}[(-2) \sin 12 x \sin (-6 x)] \\
=\sin 12 x \sin 6 x=(2 \sin 6 x \cos 6 x) \sin 6 x \\
=2 \sin ^{2} 6 x \cos 6 x
\end{array}$
Asked in: MHT CET 2020 (13 Oct Shift 1)
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